The ratio of nucleus density of nitrogen nucleus \( ({}_{7}^{14}\mathrm{N}) \) to the nucleus density of oxygen nucleus \( ({}_{8}^{16}\mathrm{O}) \) will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1 : 1
The radius of a nucleus of mass number A is \( R = R_0A^{1/3} \), so its volume is proportional to \( R^{3} \), that is, to A. Its mass is also proportional to A (each nucleon has a mass of about 1 u). So the nuclear density does not depend on A.
\( \displaystyle \rho = \frac{\text{mass}}{\text{volume}} = \frac{A\,m_{\text{nucleon}}}{\frac{4}{3}\pi R_0^{3}A} = \frac{3m_{\text{nucleon}}}{4\pi R_0^{3}} \)
This is the same for nitrogen (A = 14) and oxygen (A = 16), so \( \rho_N : \rho_O = 1 : 1 \).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Size of the Nucleus – “… It has been found that a nucleus of mass number A has a radius
\( \displaystyle R = R_0 A^{1/3} \qquad (13.5) \)
where \( R_0 = 1.2 \times 10^{-15} \) m (= 1.2 fm; 1 fm = \( 10^{-15} \) m). This means the volume of the nucleus, which is proportional to \( R^3 \) is proportional to A. Thus the density of nucleus is a constant, independent of A, for all nuclei.”